Question:
In a children-park an inclined plane is constructed with an angle of incline $45^{\circ}$ in the middle part. Find the acceleration of a boy sliding on it if the friction coefficient between the cloth of the boy and the incline is $0.6$ and $\mathrm{g}=10 \mathrm{~m} / \mathrm{s}^{2}$
.
Solution:
$N=m g \cos 45^{\circ}$ ........(perpendicular to incline)
$m g \sin 45-f f=m a$ .(along the incline)
$m g \sin 45-\mu N=m a$
$g \sin 45-\mu g \cos 45=a$
$a=2 \sqrt{2} \mathrm{~m} / \mathrm{s}^{2}$
$10 \times \frac{1}{\sqrt{2}}-0.6 \times 10 \times \frac{1}{\sqrt{2}}=a$
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